1000 e^(kt) = 1274
e^(5k) = 1.274
take ln of both sides
5k = ln 1.274
k = ln 1.274/5
= .048323...
= 4.83 % rounded to the nearest hundreth
Suppose that the number of bacteria in a certain population increases according to a continuous exponential growth model. A sample of 1000
bacteria selected from this population reached the size of 1274
bacteria in five hours. Find the hourly growth rate parameter.
Note: This is a continuous exponential growth model.
Write your answer as a percentage. Do not round any intermediate computations, and round your percentage to the nearest hundredth.
1 answer